Electronics-I Lecture 24

# Electronics-I Lecture 24 - Page421,Exercise5.21

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Assignment 5 Page 421 , Exercise 5.21 Page 426 , Exercise D5.22, D5.23, 5.24  Page 427 , Exercise D5.25, D5.26 Page 523 , Problem 5.67 (a,b) Page 524 , Problem 5.69  Page 525 , Problem 5.74, 5.79 (a,b,c,d,e) Submission date : Wednesday 13 th  Oct 2010 Quiz No.5    Next Week Quiz  No 5

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T.A Name: Waqar Anjum  Roll No.06-0452 Tel: 0321-5040161 Email: [email protected]
Example 5.4  2200 β  = 100,                   v BE   = 0.7 V given V E  = 4 – 0.7 = 3.3 V I E  = (3.3 – 0)/3.3              = 1mA I C  = I E α  = 0.99 mA α = β /( β +1) = 0.99 V C  = 10 – I C R C  = 5.3 V I B  = I E /( β +1) = I C / β         = 0.01 mA I B  = 0.01 mA 3 Electronics I Lecture 24

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2200 β  = 50,100, v BE  = 0.7 V given V E  = 6 – 0.7 = 5.3 V I E  = (5.3 – 0)/3.3 = 1.6mA 2200 α = β /( β +1) =0.98, 0.99  I C
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## This note was uploaded on 12/12/2010 for the course EE 1 taught by Professor Shahidanwar during the Spring '10 term at NUCES.

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Electronics-I Lecture 24 - Page421,Exercise5.21

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